Werner's theory of coordination compounds: Worked examples

Everything we use was defined in the parent. Two words we lean on constantly, restated in plain English:

Before any example, here is the picture we are building in every case — the metal at the centre, ligands on fixed spatial arms (secondary valency), counter ions floating free (primary valency):

Figure — Werner's theory of coordination compounds

How to read it: the blue centre is the metal; the yellow arms are the six fixed secondary-valency positions (octahedron, CN 6) that grip ligands; the red ions drifting outside are counter ions held only by primary valency — those are the ones that swim away in water and get caught by $A g N O_{3}$ .

The one formula we reuse in every example:

The scenario matrix

Every Werner problem is one of these cells. Think of the columns as "knobs": how many ligands are neutral, how many are negative, whether any counter ion is a cation ( $K^{+}$ , etc.), and what the resulting complex charge turns out to be.

Cell	Situation class	Sign of complex charge	Ionizable ions	Example
A	All ligands neutral (only $N H_{3}$ / $H_{2} O$ ), anion counter ions	positive (= oxidation state)	many anions	Ex 1
B	Mixed ligands (some neutral, some $- 1$ )	positive but reduced	fewer anions	Ex 2
C	All ligand slots filled by anions → neutral complex	zero	none (non-electrolyte)	Ex 3
D	Anion counter ions ⇒ cation counter ions ( $K^{+}$ )	negative	cations ionize	Ex 4
E	Degenerate / zero test: predict $A g N O_{3}$ precipitate = 0	any	check the free $C l^{-}$ only	Ex 3 & 5
F	Polydentate trap: CN ≠ number of molecules	positive	anions	Ex 6
G	Real-world word problem: reverse-engineer formula from lab data	inferred	inferred	Ex 7
H	Exam twist: two ions look identical, conductivity distinguishes them	equal magnitude, different structure	same count, different chemistry	Ex 8
I	CN = 4 (Pt²⁺ / Cu²⁺): the smaller secondary valency	positive or negative	depends	Ex 9
J	Zero-oxidation-state metal (e.g. $N i (C O)_{4}$ ): neutral, non-ionic	zero	none	Ex 10

We now hit every cell.

Worked example

Ex 1 — Cell A. $C o C l_{3} \cdot 6 N H_{3}$ : all-neutral ligands. Statement: Deduce the Werner formula, complex charge, ions in solution, and $C l^{-}$ precipitated by $A g N O_{3}$ .

Forecast: (cover the answer, guess) — How many $C l^{-}$ do you expect to precipitate? Is the complex charge $+ 1, + 2$ , or $+ 3$ ?

Oxidation state of Co. $N H_{3}$ is neutral ( $0$ ), each $C l$ is $- 1$ , whole compound neutral: $Co + 3 (- 1) + 6 (0) = 0 \Rightarrow Co = + 3$ . Why this step? We must never assume $+ 3$ — we read it off the raw formula first.
CN of Co(III) is fixed at 6. Why? Werner postulate 2 — the secondary valency is fixed regardless of what's dissolved.
All 6 secondary slots are taken by $N H_{3}$ . Why? We have exactly six $N H_{3}$ , each a neutral monodentate ligand (one donor atom, one bond). No room and no need for $C l^{-}$ inside.
Complex charge: $Q = (+ 3) + 6 \times (0) = + 3$ . Why? $N H_{3}$ is neutral, so it contributes $0$ ; the metal's $+ 3$ stands alone.
Counter ions: need $3 C l^{-}$ outside to cancel $+ 3$ . Formula $[C o (N H_{3})_{6}] C l_{3}$ . Why this step? The whole compound must be electrically neutral, so the outside charge must be equal and opposite to the $+ 3$ complex; three singly-negative $C l^{-}$ are the only anion available.
Ions in solution: $[C o (N H_{3})_{6}]^{3 +} + 3 C l^{-} = 4$ particles. Why this step? In water the primary-valency bonds break, so the compound splits into one intact complex ion plus each freed counter ion as a separate particle — that particle count is what conductivity measures.

Verify: All 3 $C l^{-}$ are outside ⇒ all ionize ⇒ 3 precipitate with $A g N O_{3}$ . Charge check: $+ 3$ cancels $3 \times (- 1) = - 3$ . ✓

Worked example

Ex 2 — Cell B. $C o C l_{3} \cdot 4 N H_{3}$ : mixed ligands. Statement: Only 1 $C l^{-}$ precipitates with $A g N O_{3}$ . Find the formula and confirm the ion count.

Forecast: If 4 slots are $N H_{3}$ but CN is 6, how many $C l^{-}$ must move inside? What complex charge does that give?

Oxidation state of Co. $Co + 3 (- 1) + 4 (0) = 0 \Rightarrow Co = + 3$ . Why this step? Derive $+ 3$ from the raw formula, don't assume it.
CN = 6, but only 4 $N H_{3}$ . Why? Six slots must be filled; four are taken ⇒ two slots empty.
Fill the 2 empty slots with $C l^{-}$ . Why? The lab says only 1 $C l^{-}$ is free (precipitates). So $3 - 1 = 2$ chlorides are locked inside as ligands.
Complex charge: $Q = (+ 3) + 4 (0) + 2 (- 1) = + 3 - 2 = + 1$ . Why? Each inside $C l^{-}$ carries $- 1$ and cancels part of the metal's charge.
Counter ions: one $C l^{-}$ outside cancels $+ 1$ . Formula $[C o (N H_{3})_{4} C l_{2}] C l$ . Why this step? Neutrality demands the outside charge cancel the $+ 1$ complex, so exactly one $C l^{-}$ ( $- 1$ ) sits outside.
Ions in solution: $[C o (N H_{3})_{4} C l_{2}]^{+} + C l^{-} = 2$ particles. Why this step? Only the primary-valency $C l^{-}$ breaks free; the complex ion stays whole ⇒ 2 particles total.

Verify: Free $C l^{-}$ = 1 ⇒ 1 AgCl precipitate, matches the given clue. Charge: $+ 1 + (- 1) = 0$ . ✓

Notice how moving Cl inside both lowers the complex charge and hides it from $A g N O_{3}$ — the key mechanism of the whole series.

Worked example

Ex 3 — Cells C & E. $C o C l_{3} \cdot 3 N H_{3}$ : neutral complex, zero test. Statement: Predict the conductivity and $A g N O_{3}$ result.

Forecast: This is the "degenerate" case. Guess: how many ions in solution? How many $C l^{-}$ precipitate?

Oxidation state of Co. $Co + 3 (- 1) + 3 (0) = 0 \Rightarrow Co = + 3$ . Why this step? Confirm $+ 3$ from the raw formula.
CN = 6, only 3 $N H_{3}$ ⇒ 3 empty slots. Why? Same fixed-CN logic.
All 3 $C l^{-}$ go inside to fill the slots. Why? There is no other ligand available; every chloride is now a ligand.
Complex charge: $Q = (+ 3) + 3 (0) + 3 (- 1) = 0$ . Why? The three inside $C l^{-}$ fully cancel the metal's $+ 3$ .
Counter ions: zero needed. Formula $[C o (N H_{3})_{3} C l_{3}]$ — a non-electrolyte. Why this step? A complex with charge $0$ needs no outside ion to balance it, so there is simply nothing outside the brackets.
Ions in solution: $0$ separate ions — the whole neutral molecule stays intact. Why this step? With no primary-valency bonds to break, water cannot split it; hence conductivity ≈ 0.

Verify (Cell E, the zero test): No ions form ⇒ conductivity ≈ 0 (see Conductivity and ionization of electrolytes); 0 free $C l^{-}$ ⇒ 0 AgCl precipitate. This flat-zero reading was Werner's smoking gun. ✓

The falling conductivity across the whole series is the clean visual:

How to read it: the x-axis lists the four members of the series as more $N H_{3}$ is replaced by inside- $C l^{-}$ ; the left y-axis (blue bars) counts total ions in solution and the right y-axis / yellow bars count free $C l^{-}$ (= AgCl precipitated). Both fall from left to right and hit zero at $[C o (N H_{3})_{3} C l_{3}]$ — the red arrow marks Werner's "conductivity collapse" that proved chloride had migrated inside the coordination sphere.

Worked example

Ex 4 — Cell D. $P tC l_{4} \cdot 2 K C l$ : cation counter ions, negative complex. Statement: Pt(IV) has CN 6. Find the formula and how many $C l^{-}$ precipitate with $A g N O_{3}$ .

Forecast: Adding $2 K C l$ gives extra $C l^{-}$ and $K^{+}$ . Will the complex be positive or negative this time?

Oxidation state of Pt. $K$ is $+ 1$ , each $C l$ is $- 1$ : $Pt + 6 (- 1) + 2 (+ 1) = 0 \Rightarrow Pt = + 4$ . Why this step? Read $+ 4$ off the raw combined formula, don't assume it.
Total available $C l^{-}$ : $4$ (from $P tC l_{4}$ ) $+ 2$ (from $2 K C l$ ) $= 6$ . Why count both? All chlorides are candidate ligands.
Fill all 6 CN slots with $C l^{-}$ . Why? There are exactly 6 chlorides and 6 slots — a perfect fit, no $N H_{3}$ present.
Complex charge: $Q = (+ 4) + 6 (- 1) = - 2$ . Why? Six $- 1$ ligands overshoot the $+ 4$ metal by $2$ , giving a negative complex $[P tC l_{6}]^{2 -}$ .
Counter ions: to cancel $- 2$ we need cations: $2 K^{+}$ . Formula $K_{2} [P tC l_{6}]$ . Why this step? Neutrality now requires positive counter ions; the two $K^{+}$ from the added $K C l$ supply exactly $+ 2$ to cancel the $- 2$ complex.
Ions in solution: $2 K^{+} + [P tC l_{6}]^{2 -} = 3$ particles. Why this step? The primary-valency bonds break the two $K^{+}$ free while the anionic complex stays intact ⇒ 3 particles; note the ionizing species are now cations.

Verify: Every $C l^{-}$ is a ligand (inside) ⇒ 0 free $C l^{-}$ ⇒ 0 AgCl. The ionizing species are now $K^{+}$ , not $C l^{-}$ . ✓ This is the sign-flip case people forget.

Worked example

Ex 5 — Cell E variant. $[C r (H_{2} O)_{6}] C l_{3}$ vs $[C r (H_{2} O)_{4} C l_{2}] C l \cdot 2 H_{2} O$ : the "same formula, different free Cl" twist. Statement: Both are $C r C l_{3} \cdot 6 H_{2} O$ by weight. One precipitates 3 $C l^{-}$ , the other 1. Explain.

Forecast: Same overall formula — so what physically differs?

Oxidation state of Cr. $H_{2} O$ is neutral, each $C l$ is $- 1$ : $Cr + 3 (- 1) + 6 (0) = 0 \Rightarrow Cr = + 3$ . Why this step? Derive $+ 3$ from the raw formula first.
Cr(III) CN = 6 in both. Why? Fixed secondary valency.
Violet form: all 6 slots are $H_{2} O$ (neutral). $Q = (+ 3) + 6 (0) = + 3$ ⇒ $[C r (H_{2} O)_{6}] C l_{3}$ , 3 free $C l^{-}$ . Why 3 free? All three $C l^{-}$ are counter ions outside ⇒ all ionize.
Green form: 4 slots $H_{2} O$ , 2 slots $C l^{-}$ inside; the other 2 waters are water of crystallisation (see the definition above — trapped in the lattice, not bonded to the metal, so they do not change the ion count). $Q = (+ 3) + 4 (0) + 2 (- 1) = + 1$ ⇒ $[C r (H_{2} O)_{4} C l_{2}] C l \cdot 2 H_{2} O$ , 1 free $C l^{-}$ . Why 1 free? Only the single outside $C l^{-}$ ionizes; the two inside are ligands, and the two lattice waters do nothing to conductivity or precipitate.

Verify: Same molecular formula, but $A g N O_{3}$ gives 3 vs 1 AgCl — exactly the kind of hydrate/ionization isomerism the coordination sphere predicts. Charge checks: $+ 3 - 3 = 0$ and $+ 1 - 1 = 0$ . ✓ (Related: Isomerism in coordination compounds (cis-trans, optical).)

Worked example

Ex 6 — Cell F. Polydentate trap: $C o C l_{3} \cdot 3 (e n)$ , where en = ethylenediamine. Statement: en is bidentate (two donor N atoms per molecule). Find the formula.

Forecast: Only 3 molecules of en — is CN therefore 3? (This is the trap.)

Oxidation state of Co. en is neutral, each $C l$ is $- 1$ : $Co + 3 (- 1) + 3 (0) = 0 \Rightarrow Co = + 3$ . Why this step? Read $+ 3$ off the raw formula.
en binds at 2 points. Why this matters? CN counts donor atoms, not molecules. $3$ en $\times 2 = 6$ donor atoms.
CN = 6 is satisfied by 3 en alone. Why? $6$ matches Co(III)'s fixed CN exactly ⇒ no $C l^{-}$ needed inside.
Complex charge: en is neutral, so $Q = (+ 3) + 3 \times (0) = + 3$ ⇒ $[C o (e n)_{3}]^{3 +}$ .
Counter ions: $3 C l^{-}$ outside. Formula $[C o (e n)_{3}] C l_{3}$ . Why this step? The $+ 3$ complex needs $- 3$ outside to be neutral ⇒ three $C l^{-}$ .
Ions in solution: $[C o (e n)_{3}]^{3 +} + 3 C l^{-} = 4$ particles. Why this step? All three counter-ion $C l^{-}$ break free; the chelate complex stays whole.

Verify: 3 $C l^{-}$ precipitate. If you'd wrongly set CN = 3, you'd predict the wrong geometry and isomer count. ✓ See Ligands — classification (mono/poly-dentate, chelate).

Worked example

Ex 7 — Cell G. Real-world word problem: reverse-engineer from lab data. Statement: A cobalt compound of composition $C o C l_{3} \cdot 5 N H_{3}$ dissolves to give a solution whose conductivity matches a salt yielding 3 ions, and $A g N O_{3}$ precipitates 2 mol $C l^{-}$ per mol. Reconstruct the coordination sphere.

Forecast: From "3 ions" and "2 Cl precipitate" alone, can you place every atom before writing brackets?

Oxidation state of Co. $Co + 3 (- 1) + 5 (0) = 0 \Rightarrow Co = + 3$ . Why this step? Derive $+ 3$ before touching structure.
2 free $C l^{-}$ ⇒ 1 $C l^{-}$ is locked inside. Why? $3 - 2 = 1$ chloride is not detected by $A g N O_{3}$ , so it is a ligand.
Fill CN 6: 5 $N H_{3}$ + 1 $C l^{-}$ inside $= 6$ donor atoms. ✓ Why? Fixed CN must be reached.
Complex charge: $Q = (+ 3) + 5 (0) + 1 (- 1) = + 2$ ⇒ $[C o (N H_{3})_{5} C l]^{2 +}$ .
Counter ions: $2 C l^{-}$ outside. Formula $[C o (N H_{3})_{5} C l] C l_{2}$ . Why this step? The $+ 2$ complex needs $- 2$ outside ⇒ two $C l^{-}$ , which is exactly the two the lab saw precipitate.
Ions in solution: $[C o (N H_{3})_{5} C l]^{2 +} + 2 C l^{-} = 3$ particles. Why this step? Two counter-ion chlorides free themselves; the complex stays whole ⇒ 3, matching the conductivity clue.

Verify: Predicted ions $= 3$ ✓ (matches conductivity). Predicted precipitate $= 2$ ✓ (matches $A g N O_{3}$ ). Both lab numbers reproduced from the structure — that's how Werner actually worked. See IUPAC nomenclature of coordination compounds for naming it (pentaamminechloridocobalt(III) chloride).

Worked example

Ex 8 — Cell H. Exam twist: distinguish $[C o (N H_{3})_{6}] C l_{3}$ from $[C r (N H_{3})_{5} C l] C l_{2} \cdot N H_{3}$ by conductivity alone. Statement: Two salts weigh the same per formula and both contain 3 Cl. Which conducts more, and why?

Forecast: More ions ⇒ more conductivity. Guess which one wins before working it out.

Oxidation states. Salt 1: $Co + 3 (- 1) + 6 (0) = 0 \Rightarrow + 3$ . Salt 2: $Cr + 3 (- 1) + 6 (0) = 0 \Rightarrow + 3$ (the extra $N H_{3}$ is neutral). Why this step? Both metals are $+ 3$ , so any difference must come from structure, not charge.
Salt 1 structure: all 3 Cl outside ⇒ $[C o (N H_{3})_{6}] C l_{3} \to [C o (N H_{3})_{6}]^{3 +} + 3 C l^{-} =$ 4 ions. Why? All-neutral ligands (Cell A), so every Cl is a counter ion.
Salt 2 structure: one Cl is a ligand inside (CN 6 = 5 $N H_{3}$ + 1 $C l$ ), so ⇒ $[C r (N H_{3})_{5} C l] C l_{2} \to [C r (N H_{3})_{5} C l]^{2 +} + 2 C l^{-} =$ 3 ions. The sixth $N H_{3}$ is a molecule of crystallisation trapped in the lattice — not bonded to the metal — so it does not raise the ion count. Why? Same logic as water of crystallisation: unbound, uncharged, invisible to conductivity.
Rank conductivity: more ions ⇒ higher molar conductivity ⇒ Salt 1 (4 ions) > Salt 2 (3 ions). Why this step? Molar conductivity rises with the number of free ions per formula unit.

Verify: Independent second test — $A g N O_{3}$ precipitate: Salt 1 gives 3, Salt 2 gives 2, agreeing with the ion ranking. Two measurements, one structure. ✓

Worked example

Ex 9 — Cell I. Coordination number 4: $P tC l_{2} \cdot 2 N H_{3}$ (Pt²⁺). Statement: Pt(II) has fixed secondary valency 4. Find the formula, ion count and $A g N O_{3}$ result.

Forecast: With CN 4 (not 6!), and 2 $N H_{3}$ present, how many $C l^{-}$ go inside? Positive, negative or neutral complex?

Oxidation state of Pt. $N H_{3}$ neutral, each $C l$ is $- 1$ : $Pt + 2 (- 1) + 2 (0) = 0 \Rightarrow Pt = + 2$ . Why this step? Derive $+ 2$ from the raw formula; this is what makes CN 4 the right choice.
CN of Pt(II) is fixed at 4. Why? Different metals have different fixed secondary valency — Pt²⁺, Cu²⁺ → 4 (see Coordination number and geometry). Using 6 here is the classic slip.
Fill the 4 slots: 2 $N H_{3}$ + 2 $C l^{-}$ inside $= 4$ donor atoms. Why? Only 2 neutral $N H_{3}$ are available, so the remaining 2 slots must be the two chlorides.
Complex charge: $Q = (+ 2) + 2 (0) + 2 (- 1) = 0$ . Why? Two $- 1$ ligands exactly cancel the $+ 2$ metal ⇒ a neutral square-planar complex $[P t (N H_{3})_{2} C l_{2}]$ .
Counter ions: none needed. Why this step? Charge $0$ ⇒ nothing sits outside the brackets.
Ions in solution: $0$ — a non-electrolyte. Why this step? No primary-valency bonds to break ⇒ conductivity ≈ 0.

Verify: All $C l^{-}$ are inside ⇒ 0 AgCl precipitate; conductivity ≈ 0. This neutral $[P t (N H_{3})_{2} C l_{2}]$ is exactly the cis/trans pair (cisplatin!) — see Isomerism in coordination compounds (cis-trans, optical). Charge: $+ 2 - 2 = 0$ . ✓

The CN-4 square-planar shape (contrast with the CN-6 octahedron above) is worth seeing:

How to read it: the blue metal sits at the centre of a square with four fixed positions; here two hold $N H_{3}$ (yellow) and two hold $C l$ (red) — all inside the coordination sphere, so nothing floats free and $A g N O_{3}$ sees zero chloride. The two arrangements (adjacent = cis, opposite = trans) are drawn to show why a neutral CN-4 complex still has isomers.

Worked example

Ex 10 — Cell J. Zero oxidation state: $N i (C O)_{4}$ . Statement: $C O$ is a neutral ligand and nickel here has oxidation state 0. Find the complex charge, ion count and $A g N O_{3}$ result.

Forecast: With a zero-charge metal and neutral ligands, what could the complex charge possibly be?

Oxidation state of Ni. $C O$ is neutral ( $0$ ), and the whole molecule is neutral: $Ni + 4 (0) = 0 \Rightarrow Ni = 0$ . Why this step? We derive the unusual $0$ rather than assume it — proof that the metal need not carry charge.
CN = 4, all four slots are $C O$ . Why? Four neutral $C O$ donor atoms fill the secondary valency; no anion is present or needed.
Complex charge: $Q = (0) + 4 (0) = 0$ . Why? Zero metal charge plus zero ligand charge = zero.
Counter ions: none. Why this step? A neutral complex needs no balancing ion, so there is nothing outside the brackets.
Ions in solution: $0$ — a molecular, non-ionic compound. Why this step? No primary-valency bonds exist to break, so it does not ionize at all.

Verify: No $C l$ present at all ⇒ 0 AgCl; conductivity ≈ 0. Charge: $0 + 0 = 0$ . ✓ This is the pure "everything is zero" corner of the matrix.

Recall Self-test: place the case before you solve

$[N i (C O)_{4}]$ — which cell? ::: Cell J (zero oxidation state, all-neutral CO ligands ⇒ neutral complex, non-electrolyte, 0 ions). $K_{3} [F e (C N)_{6}]$ — which cell? ::: Cell D (negative complex $[F e (C N)_{6}]^{3 -}$ , cation counter ions $3 K^{+}$ , 4 ions, 0 free $C N^{-}$ from the sphere). $[P t (N H_{3})_{2} C l_{2}]$ — which cell? ::: Cell I (CN = 4, neutral complex, 0 ions, 0 AgCl). Why does $[C o (e n)_{3}] C l_{3}$ trap students? ::: Cell F — en is bidentate, so 3 molecules give CN 6, not CN 3. Compound gives 2 ions and 1 $C l^{-}$ precipitate from $C o C l_{3} \cdot 4 N H_{3}$ — formula? ::: $[C o (N H_{3})_{4} C l_{2}] C l$ (Cell B).